Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=1010;
int n,tot,vis[maxn*10];
ll prime[maxn],ans[maxn][maxn];
void init() {
	int tmp;
	for (int i=2;i<=8000;i++) {
		if (!vis[i]) prime[++tot]=i;
		if (tot==1000) break;
		for (int j=1;j<=tot;j++) {
			tmp=i*prime[j];
			if (tmp>=maxn*10) break;
			vis[tmp]=1;
			if (i%prime[j]==0) break;
		}
	}
}
ll gcd(ll x,ll y) {
	if (!x||!y) return x+y;
	return gcd(y,x%y);
}
ll lcm(ll x,ll y) {
	return x/gcd(x,y)*y;
}
int main() {
	//freopen("1.txt","r",stdin);
	init();
	read(n);
	if (n==2) {
		printf("4 7\n23 10\n");
		return 0;
	}
	for (int i=0;i<=n+1;i++)
		ans[0][i]=ans[i][0]=ans[n+1][i]=ans[i][n+1]=1;
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
		if ((i+j)%2==0)
			ans[i][j]=prime[(i+j)/2]*prime[n+(i-j)/2+(n+1)/2];
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++)
		if ((i+j)%2==1)
			ans[i][j]=lcm(lcm(ans[i-1][j],ans[i][j-1]),lcm(ans[i+1][j],ans[i][j+1]))+1;
	for (int i=1;i<=n;i++) {
		for (int j=1;j<=n;j++)
			printf("%lld ",ans[i][j]);
		printf("\n");
	}
	return 0;
}